Proving Differentiability of f(x,y) at (0,0)

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Ok, so I have f(x,y)=(p(x)+q(y))/(x^2+y^2) where (x,y)NOT=0 and f(0,0)=0. the basic idea of the function is that the numerator contains 2 polynomials>2nd order. and the denominator has a Xsquared+ysquared. I have to prove that if f(x,y) is differentiable at (0,0) then its partial derivatives fx and fy are both continous. I need something rigorous i was thinking of doing something where the tangent h(x,y) approximates f(x,y) as dx, dy go to 0. and then from there... IDK i need all help i can get. Thank you in advance
 
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What is the definition of "differentiable" at a point, for functions of several variables?
 


that the partial derivatives are continuous in a neighborhood BUT that proves that if partial derivs are cont., then the function is differentiable. I need to prove that if its differentiable, the partials are continous
 


Then I ask again, "What is the definition of "differentiable" at a point, for functions of several variables?:"

A definition is always an "if and only if" statement so what you give is clearly NOT a definition.
 


hmmm. A function is differentiable at a point if and only if Fx and Fy are continous? i still don't know where to start the proof for the generic problem though.
 
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In part B of this question i proved that if Fx and Fy are continuous then f(x,y) is differentiable becuase a tangent plane exists there and i used lim at (0,0) (h(x,y)-f(x,y))/sqroot(x^2+Y^2) goes to 0 to complete the proof that it is differentiable. I don't know why I can't go the other way.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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